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Find the Cartesian equation of the locus of the point $P$ which is the intersection of the tangent lines at a point $t$ on the graphs of $y=e^x$ and $y=\ln(x)$

My attempt

Equation of tangent at point $(t,e^t)$ on $y=e^x$

$y=e^tx+e^t(1-t)$

Equation of tangent at point $(t,\ln(t))$ on $y=\ln(x)$

$y=\frac{1}{t}x+\ln(t)-1$

So the intersection of the two lines is when

$(e^t-\frac{1}{t})x=\ln(t)-1-e^t(1-t)$

So the x coordinate of P is

$x=\frac{t\ln(t)-t+te^t-t^2e^t}{te^t-1}$

I plug this into one of the tangent line equations and I get a really messy value for the y coordinate of P.

I suspect that there is no Cartesian equation of the locus and that it has to be expressed parametrically but I am not sure since it is a question in my textbook. Is there another way to tackle this problem? Some help would be greatly appreciated.

Amrit Cassim
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  • If i am not mistaken, the equation for $y$ is $y=\frac{\ln(t)+\frac{x}{t}-1+xe^t+e^t(1-t)}{2}$ – Hushus46 Aug 18 '17 at 05:24
  • Also I'm not sure how simple an analytic solution to this is, because the term $f(t)=te^t$ in your last equation has solutions only to be known as the Lambert W function. Give it a look, maybe you will find something key to solving this. – Hushus46 Aug 18 '17 at 05:34

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